Can someone explain me how to solve this exercise? **Chemical Reaction Problem**Chemical Equation:\[ \text{H}_2\text{SO}_4(\text{aq}) + \text{Pb(C}_2\text{H}_3\text{O}_2)_2(\text{aq}) \rightarrow \text{PbSO}_4(\text{s}) + 2 \text{HC}_2\text{H}_3\text{O}_2(\text{aq}) \]Aqueous solutions of sulfuric acid and lead (II) acetate react to form solid lead (II) sulfate and a solution of acetic acid. If 16.5 g of sulfuric acid and 17.2 g of lead (II) acetate are mixed, calculate the number of grams of sulfuric acid, lead (II) acetate, lead (II) sulfate, and acetic acid present in the mixture after the reaction is complete.**Reaction Analysis Questions:**1. Which chemical is the limiting reactant? - [Dropdown to select answer]2. How many grams of sulfuric acid are leftover? - [Input box] grams3. How many grams of lead (II) acetate are leftover? - [Input box] grams4. How many grams of lead (II) sulfate are produced? - [Input box] grams5. How many grams of acetic acid are produced? - [Input box] grams**Note:** This problem requires understanding of stoichiometry principles to determine the limiting reactant and calculate the products and leftovers of the reactant.

A reaction between H2SO4 and Lead acetate, which gives Lead sulphate and acetic acid.The balanced reaction is H2SO4(aq) + Pb(C2H3O2)2(aq) → PbSO4(s) + 2HC2H3O2(aq) To find the no.of moles Mass of H2SO4 is 16.5 g Mass of Pb(C2H3O2)2 is 17.2 g No.of moles of H2SO4 = Mass of H2SO4Molar mass of H2SO4 = 16.5 g98.079 g/mol = 0.1682 mol No.of moles of Lead acetate = Mass of Lead acetateMolar mass of Lead acetate = 17.2 g325.29 g/mol = 0.05288 mol To find the theoretical yield Yield of PbSO4 due to H2SO4 = Mass of H2SO4Molar mass of H2SO4*Molar mass of PbSO4 = 16.5 g98.079 g/mol*303.26 g/mol = 51.018 g Yield of PbSO4 due to Pb(C2H3O2)2 =Mass of Pb(C2H3O2)2Molar mass of Pb(C2H3O2)2*Molar mass of PbSO4 = 17.2 g325.29 g/mol*303.26 g/mol = 16.035 g The limiting reagent Pb(C2H3O2)2 is the limiting agent, as it gives the lowest yield. As per the balanced equation, 1 mole of H2SO4 reacts with 1 mole of Lead acetate to give 1 mole of PbSO4 and 2 moles of acetic acid. 0.05288 mol of Lead acetate would react with 0.05288 mol of H2SO4 The remaining H2SO4 is 0.11532 mol To find the mass of H2SO4 left over Mass of H2SO4 = No.of moles * molar mass = 0.11532 mol * 98.079 g/mol = 11.31 g To find the mass of Acetic acid produced Yield of acetic acid due to lead acetate = Mass of Lead acetateMolar mass of Lead acetate*21* Molar mass of acetic acid = 17.2 g325.29 g/mol *2*60.052 g/mol = 6.35 gLimiting reactant is Pb(C2H3O2)2 H2SO4 left over is 11.31 g Since, Pb(C2H3O2)2 is the limiting reagent 0.0 g of Pb(C2H3O2)2 is left over. PbSO4 produced is 16.035 g CH3COOH produced is 6.35 g